Prove these formulas for the distance between and a line/plane.

[Jow]

I have two problems that are very similar.
Problem 1: Prove that in the case where the line (L) is in ℝ2 and its equation has the general form ax+by=c, the distance from point B=(x$_0{}$,y$_0{}$) to the line d(B,L) is given by the first formula.

Problem 2: Prove that, in general, the distance d(B, P) from the point B=(x$_0{}$,y$_0{}$,z$_0{}$) to the plane whose general equation is ax+by+cz=d is given by the second formula.



2. Formula 1: d(B,L)=$\frac{\left|ax_0{}+by_0{}-c\right|}{\sqrt{a^2+b^2}}$

Formula 2: d(B,L)=$\frac{\left|ax_0{}+by_0{}+cz_0{}-d\right|}{\sqrt{a^2+b^2+c^2}}$

3. I don't even know where to begin with these.

 

[Dick]

I have two problems that are very similar.
Problem 1: Prove that in the case where the line (L) is in ℝ2 and its equation has the general form ax+by=c, the distance from point B=(x$_0{}$,y$_0{}$) to the line d(B,L) is given by the first formula.

Problem 2: Prove that, in general, the distance d(B, P) from the point B=(x$_0{}$,y$_0{}$,z$_0{}$) to the plane whose general equation is ax+by+cz=d is given by the second formula.



2. Formula 1: d(B,L)=$\frac{\left|ax_0{}+by_0{}-c\right|}{\sqrt{a^2+b^2}}$

Formula 2: d(B,L)=$\frac{\left|ax_0{}+by_0{}+cz_0{}-d\right|}{\sqrt{a^2+b^2+c^2}}$

3. I don't even know where to begin with these.

Find a unit normal vector to the line. Then you can find the distance by taking the dot product of that vector with the difference between (x0,y0) and any point on the line. The same approach will work for the plane.

 

[Jow]

That makes sense but you say to use the difference between (x0, y0) and any point on the line. How would you do that? Wouldn't the point you would need to use be (x, y). Is this right?

 

[Dick]

That makes sense but you say to use the difference between (x0, y0) and any point on the line. How would you do that? Wouldn't the point you would need to use be (x, y). Is this right?

No. You want to get a specific point on the line. Put say x=0 into the equation for the line. What's the corresponding point?

 

[Jow]

It doesn't give an equation for the line. It just says to prove the formula.

 

[Dick]

It doesn't give an equation for the line. It just says to prove the formula.

It says the line is ax+by=c. If x=0 what is y?

 

[HallsofIvy]

Another way to do this problem is to use the geometric fact that the distance from a point to a line or a plane is always measured along a line perpendicular to the line or plane.

If a line is given by ax+ by= c, what is the slope of the line? What is the slope of a perpendicular line? What is the equation of a line having that slope and going through $(x_0, y_0)$?

If the plane is given by ax+ by+ cz= d, what is a vector perpendicular to that plane? What are parametric equations for a line in the direction of that vector through $(x_0, y_0, z_0)$?

 

[Jow]

I get it now. Thanks.